A bag contains $19$ unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{\mathrm{m}}{\mathrm{n}},\gcd (\mathrm{m},\mathrm{n})=1$, then ${n}^{2}-{m}^{2}$is equal to : [JEE Main 2025, 7 Apr (Shift 2)]

Question

YESPYQ · Accepted Answer

Correct answer: a) $80$. $U$: Selecting an unbiased coin. $P(U)=\frac{19}{20}$ $B$: Selecting the biased (two-headed) coin. $P(B)=\frac{1}{20}$ $H$: Getting a head on the toss. $P(H\mathrm{∣}U)=\frac{1}{2}$ (for an unbiased coin) $P(H\mathrm{∣}B)=1$ (for a two-headed coin) $P(U\mathrm{∣}H)=\frac{P(U)⋅P(H\mathrm{∣}U)}{P(U)⋅P(H\mathrm{∣}U)+P(B)⋅P(H\mathrm{∣}B)}$ $P(U\mathrm{∣}H)=\frac{\frac{19}{20}⋅\frac{1}{2}}{(\frac{19}{20}⋅\frac{1}{2})+(\frac{1}{20}⋅1)}=\frac{\frac{19}{40}}{\frac{19}{40}+\frac{2}{40}}=\frac{19}{21}$ Given $P(U\mathrm{∣}H)=\frac{m}{n}=\frac{19}{21}$, where $\text{gcd}(19,21)=1$. So, $m=19$ and $n=21$. ${n}^{2}−{m}^{2}=(n−m)(n+m)$ ${n}^{2}−{m}^{2}=(21−19)(21+19)=2\times 40=80$

A bag contains \(19\) unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head t…

Practice more JEE Maths PYQs